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vi ab = cd
=>a/b=c/d
=>a+c/b+d =a/b = c/d
=>a-c/b-d =a/b = c/d
(sgk s8 )
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\left(\frac{a+b}{c+d}\right)^3=\left(\frac{bk+b}{dk+d}\right)^3=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^3=\left(\frac{b}{d}\right)^3\left(1\right)\)
\(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}=\left(\frac{b}{d}\right)^3\left(2\right)\)
Từ (1) & (2)=>\(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(a+b+c+d\right)\left(a-b-c+d\right)\)
\(=\left(a+d\right)^2-\left(b+c\right)^2\)
\(=\left(bk+d\right)^2-\left(b+dk\right)^2\)
\(=b^2k^2+2bkd+d^2-b^2-2bkd-d^2k^2\)
\(=b^2\left(k^2-1\right)+d^2\left(1-k^2\right)\)
\(=\left(k^2-1\right)\left(b^2-d^2\right)\)(1)
\(\left(a-b+c-d\right)\left(a+b-c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=\left(bk-d\right)^2-\left(b-dk\right)^2\)
\(=b^2k^2-2bkd+d^2-b^2+2dk-d^2k^2\)
\(=k^2\left(b^2-d^2\right)-\left(b^2-d^2\right)=\left(b^2-d^2\right)\left(k^2-1\right)\)(2)
Từ (1) và (2) suy ra ĐPCM
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a^2-b^2}{ab}=\frac{\left(bk\right)^2-b^2}{bk.b}=\frac{b^2.k^2-b^2}{b^2k}=\frac{b^2\left(k^2-1\right)}{b^2k}=\frac{k^2-1}{k}\left(1\right)\)
\(\frac{c^2-d^2}{cd}=\frac{\left(dk\right)^2-d^2}{dk.d}=\frac{d^2k^2-d^2}{d^2k}=\frac{d^2\left(k^2-1\right)}{d^2.k}=\frac{k^2-1}{k}\left(2\right)\)
Từ (1) và (2)=>\(\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\).
ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}.\)
\(\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
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Ta có: \(\frac{a}{b}=\frac{c}{d}=k\)
=> a=b*k, c=d*k
Lại có: \(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\)(1)
\(\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\)(2)
Từ (1) và (2) ta suy ra:\(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)